A zoom tracking algorithm based on defocus difference

Abstract

The image clarity evaluation function was commonly used by the autofocus algorithm to evaluate the clarity of the image. And autofocus algorithms like peak search algorithm and zoom tracking algorithm are based on the evaluation value. This paper proposes an Improved Feedback Zoom Tracking method (IFZT) based on defocus difference, using the amount of defocus difference as the degree of image blur. IFZT algorithm modifies the revision criterion for feedback revision point and removes the relatively complex PID algorithm. IFZT determines the orientation of the in-focus motor position according to the amount of defocus difference and uses the depth of defocus method to estimate the ideal focus position. In this paper, the calculation formula of defocus difference and ideal focus position are deduced. Finally, the algorithm was experimented on an integrated camera; the experimental results show that: IFZT algorithm using the amount of defocus has good tracking accuracy, and has a larger promotion compared with other zoom tracking algorithms. And the overall performance of IFZT algorithm is in line with the requirements of zoom tracking algorithm.

Appendix

Derivation of formula 9 : ${I}^{{{\prime}}}\left(x,y\right)=I\left(x,y\right)+\frac{1}{4}{\sigma }^{2}{\nabla }^{2}{I}^{{{\prime}}}\left(x,y\right)$

In our application, we usually use a third-order polynomial to approximate the image.

$$I\left( {x,y} \right) = \mathop \sum \limits_{0 \le m + n \le 3} a_{m,n} x^{m} y^{n} \left\{ {\begin{array}{*{20}c} {A1 \le x \le A2} \\ {B1 \le y \le B2} \\ \end{array} } \right.,$$

(24)

where, A1, A2, B1, B2 are the image boundaries.

Using the forward S transform, we have the blurred image,

$$\begin{aligned} I^{\prime}\left( {x,y} \right) & = I\left( {x,y} \right)*h_{\sigma } \left( {x,y} \right) \\ & = \mathop \sum \limits_{0 \le m + n \le 3} \frac{{\left( { - 1} \right)^{m + n} }}{m!n!}h_{m,n} I^{m,n} , \\ \end{aligned}$$

(25)

Assuming h(x, y) is circularly symmetric, it can be shown that

$${h}_{m,n}=0{\mathrm{ for }}\left(m{\mathrm{ odd}}\right){\mathrm{or }}\left(n{\mathrm{ odd}}\right){\mathrm{ and }}{h}_{m,n}={h}_{n,m}.$$

(26)

For any point spread function,

$${h}_{{0,0}}=1.$$

(27)

From Eqs. (25) and (27), ${I}^{^{\prime}}(x,y)$ becomes

$${I}^{^{\prime}}\left(x,y\right)={I}^{{0,0}}+{\frac{1}{2!}h}_{{2,0}}{I}^{{2,0}}+{\frac{1}{2!}h}_{{0,2}}{I}^{{0,2}}.$$

(28)

From Eq. (26) we have ${h}_{{0,1}}={h}_{{1,0}}={h}_{{1,1}}=0$ and ${h}_{{0,2}}={h}_{{2,0}},$ and using inverse S transform, we can get:

$$I^{\prime } \left( {x,y} \right) = I\left( {x,y} \right) + \frac{{h_{0,2} }}{2}\left\{ {I^{\prime 2,0} \left( {x,y} \right) + I^{\prime 0,2} \left( {x,y} \right)} \right\}.$$

(29)

From the definition of moments and $\sigma$, we have ${h}_{{2,0}}={h}_{{0,2}}={\sigma }^{2}/2$, so

$$I^{\prime } \left( {x,y} \right) = I\left( {x,y} \right) + \frac{{\sigma^{2} }}{4}\left\{ {I^{\prime 2,0} \left( {x,y} \right) + I^{\prime 0,2} \left( {x,y} \right)} \right\}.$$

(30)

Note that $I^{\prime 2,0} + I^{\prime 0,2} = \nabla^{2} I^{\prime }$ corresponds to the Laplacian operation on the image $I^{\prime } \left( {x,y} \right)$, thus we can get:

$$I^{\prime } \left( {x,y} \right) = I\left( {x,y} \right) + \frac{1}{4}\sigma^{2} \nabla^{2} I^{\prime } \left( {x,y} \right).$$

(31)